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प्रश्न
Prove that : `tan"A"/(1 - cot"A") + cot"A"/(1 - tan"A") = sec"A".cosec"A" + 1`.
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उत्तर
LHS = `(sin"A"/cos"A")/(1 - cos"A"/sin"A") + (cos"A"/sin"A")/(1 - sin"A"/cos"A")`
= `(sin"A" sin"A")/(cos"A"(sin"A" - cos"A")) + (cos"A" cos"A")/((cos"A" - sin"A") sin"A"`
= `1/((sin"A" - cos"A")) [(sin^2"A")/cos"A" + (cos^2"A")/(-sin"A")]`
= `(sin^3"A" - cos^3"A")/(sin"A".cos"A"(sin"A" - cos"A"))`
= `((sin"A" - cos"A")(sin^2"A" + cos^2"A" + sin"A". cos"A"))/(sin"A". cos"A"(sin"A" - cos"A")`
= `(1 + sin"A". cos"A")/(sin"A".cos"A")`
= `1/(sin"A".cos"A") + (sin"A".cos"A")/(sin"A".cos"A")`
= `1/sin"A" . 1/cos"A"+ 1`
= sec A.cosec A + 1
= RHS
Hence proved.
संबंधित प्रश्न
If cosθ + sinθ = √2 cosθ, show that cosθ – sinθ = √2 sinθ.
Prove the following trigonometric identities.
`tan theta - cot theta = (2 sin^2 theta - 1)/(sin theta cos theta)`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove the following identities:
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (1 + cosA)/sinA`
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
If ` cot A= 4/3 and (A+ B) = 90° ` ,what is the value of tan B?
Prove the following identity :
`sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA`
Prove the following identity :
`(1 + cosA)/(1 - cosA) = (cosecA + cotA)^2`
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to
Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
